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A new characterization of the dual space of the HK-integrable functions

  • We construct the Henstock-Kurzweil (HK) integral as an extension of a linear form initially defined on L1, but which is not continuous in this space. This gives us an alternative way to prove existing results. In particular, we give a new characterization of the dual space of Henstock-Kurzweil integrable functions in terms of a quotient space.

    Citation: Juan H. Arredondo, Genaro Montaño, Francisco J. Mendoza. A new characterization of the dual space of the HK-integrable functions[J]. AIMS Mathematics, 2024, 9(4): 8250-8261. doi: 10.3934/math.2024401

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  • We construct the Henstock-Kurzweil (HK) integral as an extension of a linear form initially defined on L1, but which is not continuous in this space. This gives us an alternative way to prove existing results. In particular, we give a new characterization of the dual space of Henstock-Kurzweil integrable functions in terms of a quotient space.



    It is well known that λ is a continuous linear functional on the space of Henstock-Kurzweil integrable functions on [a,b] if there exists a function ϑ:[a,b]R of essentially bounded variation such that

    λ(f)=(HK)baϑf, (1.1)

    (see [1]). For example, in [2, Theorem 12.7] it is proved that

    infϑ1=ϑ a.eVar[ϑ1;[a,b]]2λ

    where Var[ϑ1;[a,b]] denotes the total variation of ϑ1.

    We present a new method to show that the space of essentially bounded variation functions is in fact isometric to the dual space of Henstock-Kurzweil integrable functions:

    infϑ1=ϑ a.eVar[ϑ1;[a,b]]= [ϑ1] BVm=λ,

    where [ϑ1] is an element of the quotient space, BVm, of BV with given equivalence relation. It comes out that

    A=BVm

    is an isometric isomorphism. A denotes the dual space of the Henstock-Kurzweil integrable functions normed with the Alexiewicz norm. For details, see Theorem 3 of this article. In fact, Theorem 1 of [3] shows that the functionals in the space of Henstock-Kurzweil integrable functions are of the form

    (HK)bafϑ

    where ϑ is a bounded variation function. Modification of the function ϑ on a null set gives rise to the same continuous linear functional. So, the correspondence between linear functionals and bounded variation functions is not a bijection. In this paper we show that the dual space of the Henstock-Kurzweil integrable functions is isometrically isomorphic to the space of essentially bounded variation functions seen as a quotient space.

    We mention also that this result is related to James' theorem, (see [4,5]), which assures that in a non-reflexive space like the one in question not every continuous functional reaches its supremum. But rather, there must always be values arbitrarily close to the norm of the functional.

    Furthermore, in this paper we build the Henstock-Kurzweil integral (abbreviated HK-integral), via the limit of integrals of functions that are Lebesgue integrable, thanks to the fact that L1[0,2π] is dense in the space of Henstock-Kurzweil integrable functions. In fact, on C[0,2π] we can define the positive linear form

    λ(f)=limn2π0fn(x)dx,

    with the integral in the sense of Riemann and (fn) converging to f in L1-sense. This leads to the Daniell-Stone integral and the space L1(λ) of integrable functions in this sense. It contains the Riemann integral and coincides with the usual Lebesgue space L1[0,2π], (see [6, Theorem 1.1]).

    Following a similar idea we can extend the notion of integration, inheriting naturally the properties of the previous integration theory on which it is based but adding new features. Our proofs of important results in this theory as the multiplier Theorem are elementary and easy.

    For a given fL1[0,2π] one can define the positive linear form

    m(f):=2π0f,

    which easily leads to

    |m(f)|sup0x2π|x0f|. (2.1)

    The supremum is known as the Alexiewicz norm of f, denoted as fA. The linear space L1[0,2π] with the Alexiewicz norm is normed but not complete. A basic result is the following:

    Lemma 1. The space (L1[0,2π],A) is a normed space.

    Definition 1. The completion of the space L1[0,2π] under the norm A is denoted by A.

    The map m can be extended over functions out of L1[0,2π]. The extension of the linear form m on A is denoted by the same symbol.

    Definition 2. The Henstock-Kurzweil integral is defined on A by

    (HK)2π0f(t)dt:=m(f)=limnm(fn),

    where fnf in the A-norm.

    The prefix (HK) makes it clear that it is not a Lebesgue integral.

    Remark 1. Definition 2 agrees with the classical integral of Henstock-Kurzweil (see [7]).

    The following results are well known, (see [8, Theorems 1 and 3] and [9, Theorems 2.5.1 to 2.5.12]).

    Theorem 1.

    (a) The Henstock-Kurzweil integral contains the Lebesgue integral in the sense that

    (HK)2π0f=2π0f( fL1[0,2π]).

    (b) For given f1, f2A, and βR, the equality

    (HK)2π0(f1+βf2)=(HK)2π0f1+β(HK)2π0f2

    is valid.

    (c) If fA and (fn)L1[0,2π] converges to f in A-norm, then

    fχ[0,t]:=limnfnχ[0,t]A,

    and F(t):=(HK)t0f:=(HK)2π0fχ[0,t] is continuous on [0,2π].

    Definition 3. Let v be a real valued function over R. It is said v is a bounded variation function over [0,2π] if

    Var[v,[0,2π]]:=supni=1|v(xi)v(xi1)|<

    where the supremum is taken over all partitions on [0,2π].

    The space of functions that have finite variation on [0,2π] is denoted by BV[0,2π]. In the space BV[0,2π] we introduce a norm by

    vBV:=|v(2π)|+Var[v,[0,2π]].

    Note that this norm is equivalent to the one in [10, Theorems 2.2.1 and 2.2.2]. Thus BV[0,2π] with the given norm is a Banach space.

    In order to give a characterization of the dual space (A), we use

    L1[0,2π]A  (A)L1[0,2π]=L[0,2π].

    This allows us to precisely see not only an element of (A) as an element in L[0,2π], but also as an element in the dual space of C[0,2π]. In fact, C[0,2π] is characterized by Borel measures, which in turn are related to the space of functions of bounded variation BV[0,2π], (see [11, Theorem 7.1.1] and [12, Theorem 6.3.3]).

    The multiplier theorem for functions on A is well known, but for the convenience of the reader, we give a new proof in our context.

    Theorem 2. If fA and ϑBV[0,2π], then fϑA. Its integral is given by

     (HK)2π0f ϑ=ϑ(2π)F(2π)2π0F dϑ (2.2)

    where F(t)=(HK)t0f and the integral on the right side of the equation is in the Riemann-Stieltjes sense.

    Proof. First suppose that fL1[0,2π]. Since ϑ is a bounded measurable function, then fϑL1[0,2π]. Due to [13, Theorem 3.135], for given fL1[0,2π], we can choose a sequence of functions (fn)C[0,2π] converging to f in L1-norm. If Fn(t):=t0fn, then

    FnF=sup0x2π|x0fnf|fnfL10. (2.3)

    The integration by parts formula for the Riemann-Stieltjes integral and [12, Theorem 6.2.8] imply that

    2π0ϑdFnexists due to2π0Fndϑexisting.

    Therefore, by the fundamental theorem of calculus and [14, Theorem 7.8] we get

    2π0fnϑ=2π0Fnϑ=2π0ϑdFn=ϑ(2π)Fn(2π)2π0Fndϑ. (2.4)

    From [15, Corollary H.4] we get

    |2π0Fndϑ2π0Fdϑ|FnF ϑBV0.

    This inequality and (2.3) yield, after taking limits on both sides of formula (2.4),

    2π0f ϑ=ϑ(2π)F(2π)2π0F dϑ(fL1[0,2π]). (2.5)

    On the other hand, we can define for ϑBV[0,2π] the multiplication operator

    Tϑ:L1[0,2π]A;Tϑ(f):=ϑf.

    From (2.5) we get that Tϑ is a bounded operator:

    Tϑ(f)AcfA( fL1[0,2π]), (2.6)

    where c is a positive constant. It follows from the Bounded Linear Transformation Theorem [16, Theorem 1.7] that Tϑ extends uniquely to a bounded operator from the completion of (L1[0,2π],A) to (A,A). Denoting by the same symbol the extension of the operator on A, one has

    Tϑ(f):=ϑffϑ=limnTϑ(fn)A

    whenever (fn) converges to fA. Now, using (2.5) and (2.6) and a sequence L1[0,2π]fnf in the A-norm, we get the validity of the formula for all fA. This proves the theorem.

    In the case of the classical space L1(X;dμ), with X an arbitrary measure space of positive measure μ, its dual space is the quotient space L(X;dμ)=L(X;dμ)/W, where  L(X;dμ) denotes the space of bounded measurable real functions defined in X and

    W={f:XR:f=0 a.e.}.

    The norm is defined on L(X;dμ) as:

    f=inf{M0:|f|χ[M,)W}.

    The map gFg(f)=Xfgdμ is an isometric isomorphism of L(X;dμ) onto L1(X;dμ) when (X;dμ) is σ-finite, (see [17, page 375]). We will show an analogous result for the space A. We introduce the Banach space defined by the quotient space

    BVm:=BV/Z,

    where Z:={vBV[0,2π]:v(x)=0a.e.}.

    Lemma 2. BVm is a Banach space of equivalence classes [v]:={ vBV[0,2π] : vvZ }, with the norm given by

     [v] BVm:=infvZvvBV.

    Proof. We will show that Z is a closed subspace of BV[0,2π]. Then the statement of the lemma is implied by a classical result in Functional Analysis [17, Theorem 4.2]. Given a Cauchy sequence (vn)Z in the norm of BV[0,2π] with limit v, we have that for each nN, there exists a measurable set Υn[0,2π] with

    vn|Υn0,such that γ(Υcn)=0.

    Here γ is denoting the Lebesgue measure. Therefore, due to convergence in BV[0,2π] implying in particular pointwise convergence, we get

    v(x)=limnvn(x)=0( xmNΥm).

    Furthermore,

    γ( (mNΥm)c )mγ(Υcm)=0.

    This proves that the limit v belongs to Z.

    In [18, page 241] the normalized functions are defined. For our purposes we slightly modify that concept.

    Definition 4. Let vBV[0,2π]. For x[0,2π) we define

    v+(x):=limtx+v(t),

    and v+(2π)=0. We will say that v+ is the normalization in the Alexiewicz sense of the function v. The set of all such functions will be denoted by NABV[0,2π].

    Lemma 3. Let vBV[0,2π]. Then there exists a unique element v+[v] belonging to NABV[0,2π].

    Proof. Let u,w[v] and x[0,2π). By [10, Corollary 2.1.23] the limits u+(x),w+(x) exist. Let ϵ>0 be given. There exists δ1(ϵ)>0, such that if s(x,x+δ1(ϵ)), then

    |u+(x)u(s)|<ϵ2.

    Also, there exists δ2(ϵ)>0 such that if s(x,x+δ2(ϵ)), then

    |w+(x)w(s)|<ϵ2.

    On the other hand, there necessarily exists y(x,x+δϵ) such that

    u(y)=w(y)

    where δϵ=min{δ1(ϵ),δ2(ϵ)}. So,

    |u+(x)w+(x)||u+(x)u(y)|+|w+(x)w(y)|<ϵ

    and since ϵ is arbitrary, we obtain

    u+(x)=w+(x)x[0,2π].

    Lemma 4. Given vBV[0,2π] its normalization in the Alexiewicz sense v+ belongs to BV[0,2π].

    Proof. Let w[v]. For an arbitrary partition {xj}nj=0 of [0,2π], let {xj}nj=0 such that xj<xj<xj+1 for 0jn1 and xn=xn. Given ϵ>0, we choose xj sufficiently close to xj such that

    |v+(xj)w(xj)|<ϵ2j

    for all j=0,1,...,n1. Therefore,

    nj=1|v+(xj)v+(xj1)|nj=1|v+(xj)w(xj)|+nj=1|w(xj)w(xj1)|+nj=1|v+(xj1)w(xj1)||w(2π)|+n1j=1|v+(xj)w(xj)|+nj=1|w(xj)w(xj1)|+nj=1|v+(xj1)w(xj1)||w(2π)|+n1j=1ϵ2j+nj=1|w(xj)w(xj1)|+nj=1ϵ2j1.

    Now, taking the supremum over all partitions on [0,2π] we get

    v+BV3ϵ+wBV.

    Since ϵ is arbitrary, we conclude

    v+BVwBV( w[v]).

    Corollary 1. If vBV[0,2π], then  [v] BVm=v+BV.

    Theorem 3. The spaces (A) and BVm are isometrically isomorphic:

    (a) (A)=BVm.

    (b) For every λ(A), there exists a unique [v]BVm such that

    λA= [v] BVm.

    Proof. (a) Let l(A). For given fL1[0,2π], we take

    F0(t):=t0fC0[0,2π]C[0,2π], (3.1)

    where C0[0,2π] consists of the continuous functions vanishing at zero. We define

    n(F0)=l(F0). (3.2)

    It follows that,

    |n(F0)|=|l(F0)|cF0A=cF0.

    We get a continuous linear functional defined on a subspace of C[0,2π]. From the Hahn-Banach Theorem there exists an extension on the space C[0,2π] with the same norm. We denote this extension by the same symbol n. Riesz Theorem [19, Theorem 2.14] assures that there exists a unique finite signed measure ρ so that

    n(F0)=2π0F0 dρ( F0C0[0,2π]).

    By the Jordan Decomposition [11, Theorem 5.1.8] each finite signed measure ρ is the difference of two measures μ1 and μ2, implying

    n(F0)=2π0F0(t)dμ12π0F0(t)dμ2.

    Substitution in this equation of the equality (3.1) and application of Fubini's Theorem yield

    n(F0)=2π0f(s)ds2πsdμ12π0f(s)ds2πsdμ2=2π0f(s)(μ1([s,2π]μ2([s,2π]))ds=2π0f(s)v+(s)ds. (3.3)

    Where, v(s)=μ1([s,2π])μ2([s,2π]) for all s[0,2π] is a bounded variation function [15, page 104], and v+NABV[0,2π] is the normalization of v. It yields

    l(f)=l(F0)=n(F0)=2π0f v+, (3.4)

    where

    F0(s)=s0f,( fL1[0,2π]).

    For fA, we can take a sequence (fn)L1[0,2π] such that (fn) converges to f in the A-norm. Continuity of lA together with (3.4) give

    l(f)=limn2π0v+ fn=(HK)2π0v+ f( fA). (3.5)

    Due to v+BV[0,2π] and Lemma 3, we obtain (A)BVm. The other contention is a consequence of item (a). In fact, if vBV[0,2π], then

    λ(f)=(HK)2π0fv+

    defines a continuous functional on A, and is associated with a unique element of the quotient space BVm by Lemma 3. (b) Let λ(A), and from (a) we know that there exists v+[v] representing λ. Corollary 1 and (2.2) imply

    λAv+BV= [v] BVm. (3.6)

    We now prove the reverse inequality. Let {xi}ni=1 be a partition such that

    ni=1|v+(xi)v+(xi1)| [v] BVmϵ.

    We pick up points {xi}ni=0 of [0,2π] as follows: Let x0=x0=0 and xn=xn=2π. Since v+ has a countable number of discontinuities, we can find points xi<xi<xi+1 such that v+ is continuous at every xi. Moreover, we choose each xi close enough to xi so that

    |v+(xi)v+(xi)|ϵ2i,for i=1,2,3,...,n1.

    Now we consider the partition {ξi}2n1i=0={xi}n1i=1{xi}ni=0. Note that ξ0=x0, ξ1=x1, ξ2=x1, ξ3=x2, ξ4=x2, ..., ξ2i1=xi, ξ2i=xi, ..., ξ2n3=xn1, ξ2n2=xn1, ξ2n1=xn.

    Let δ>0 such that δ<min{|ξiξi1|,i=1,...,2n1} and consider Fδ,i:[0,2π]R, to be a continuous function defined by

    Fδ,i(x)={0ifxxiorxxi+1+δ1δ(xxi)ifxi<x<xi+δ1ifxi+δxxi+111δ(xxi+1)ifxi+1<x<xi+1+δ

    for i=0,1,2,...,n2. So,

    λ(fδ,i)=(HK)2π0fδ,iv+,

    where fδ,i=Fδ,iA is defined almost everywhere. Now, it is clear that

    (HK)2π0fδ,iv+=2π0Fδ,idv+=xi+δxi(xxiδ)dv+xi+1xi+δdv+xi+1+δxi+1(1xxi+1δ)dv+.

    Using continuity in xi and right continuity in xi of v+, from [10, Corollary 2.3.4, Lemma 5.1.11] it yields

    limδ0λ(fδ,i)=[v+(xi+1)v+(xi)]

    for i=0,1,...,n2. When i=n1, we put

    Fδ,n1(x)={0ifxxn11δ(xxn1)ifxn1<x<xn1+δ1ifxn1+δx2π.

    In any case the previous argument is valid. We can suppose that each v+(xi)v+(xi1) is negative; otherwise, we change the sign of each Fδ,i in the corresponding interval, so,

    limδ0λ(fδ,i1)=|v+(xi)v+(xi1)|

    for i=1,...,n. On the other hand,

    2n1i=1|v+(ξi)v+(ξi1)|=ni=1|v+(xi)v+(xi1)|+n1i=1|v+(xi)v+(xi)|.

    Therefore,

    2n1i=1|v+(ξi)v+(ξi1)|ni=1limδ0λ(fδ,i1)+n1i=1ϵ2ilimδ0λ(fδ)+ϵ

    where fδ=ni=1fδ,i1. Thus, there exists ρ>0 such that

    |λ(fρ)|2n1i=1|v+(ξi)v+(ξi1)|2ϵni=1|v+(xi)v+(xi1)|2ϵ [v] BVm3ϵ.

    Yielding,

    λAfρA|λ(fρ)| [v] BVm3ϵ.

    It is clear that

    fρA=1.

    From this and (3.6) we conclude

    λA= [v] BVm.

    The following example shows that A is not a reflexive space. We recall that for given yA, yBV is defined by

    y(v):=2π0yv(vBV).

    A is called reflexive if every element LBVm obeys L=y for some yA.

    Example 1. Let L:BVmR be defined by

    L([v])=12( v+(2/3)v+(1/2))+12( v+((2/3))v+((1/2)) ).

    Similar arguments as given previously show that

    |L([v])| [v] BVm.

    So, this proves that LBVm. We will prove that there is no yA such that y=L. Let yA and

    Y(t)=(HK)t0y.

    We note that Y(0)=0, and suppose that Y(t) is not the identically zero function. Continuity of Y implies that we can find 0<c12π such that Y(c1)0. If 0<c1<12, we consider the function v+ equal to -1 on [0,c1) and 0 otherwise. It follows that

    y(v+)=(HK)2π0yv+=2π0Ydv+.

    By [20, Theorems 6.1.1 and 6.1.6] we have

    y(v+)=Y(c1)0=L([v]).

    In the case 1/2c12π the proof is similar. We have proved that there is no yA such that y=L.

    Corollary 2. The space A is not reflexive.

    Remark 2. A Banach space X is reflexive if and only if every continuous linear functional on the space X attains its supremum on the closed unit ball, (see [21,22]). In fact, in the proof of Theorem 3 (b) we showed that, for a given functional λ, there exists a function f such that the absolute value |λ(f)| is arbitrarily close to the norm of the functional, but does not necessarily attain this value.

    We have presented a new way of constructing the dual space of the space of integrable functions in HK, which may motivate a discussion about applying this method to similar studies in related spaces.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    Thanks to Nancy Keranen for proofreading the article. We acknowledge partial support by CONACyT-SNI.

    The authors would like to thank the anonymous referees for their careful reading and helpful suggestions that improve the presentation of this article.

    The authors declare no conflict of interest.



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